Question

(i) In figure (1), $O$ is the centre of the circle. If $\angle OAB={40}^{\circ }$ and $\angle OCB={30}^{\circ }$. Find $\angle AOC$.

(ii) In figure (2), $A,B$ and $C$ are three points on the circle with centre $O$ such that $\angle AOB={90}^{\circ }$ and $\angle AOC={110}^{\circ }$. Find $\angle BAC$.

Answer 1

(i) Join $BO$

In $△BOC$, we have:
$OC=OB$ (Radii of a circle)
$\Rightarrow \angle OBC=\angle OCB$
$\angle OBC={30}^{\circ }$ .......(i)
In $△BOA$ , we have:
$OB=OA$ (Radii of a circle)
$\Rightarrow \angle OBA=\angle OAB$ [Since, $\angle OAB={40}^{\circ }$]
$\Rightarrow \angle OBA={40}^{\circ }$........(ii)
Now, we have $\angle ABC=\angle OBC+\angle OBA$
$={30}^{\circ }+{40}^{\circ }$ [from eq (i) and (ii)]
$\therefore \angle ABC={70}^{\circ }$
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
i.e., $\angle AOC=2\angle ABC$
$=2\times {70}^{\circ }$
$\therefore \angle AOC={140}^{\circ }$

(ii)

We know that angles around a point will always add up to ${360}^{\circ }$.
So, $\angle BOC+{90}^{\circ }+{110}^{\circ }={360}^{\circ }$

$\Rightarrow \angle BOC={360}^{\circ }-({90}^{\circ }+{110}^{\circ })$
$={360}^{\circ }-{200}^{\circ }$
$\therefore \angle BOC={160}^{\circ }$
We know that $\angle BOC=2\angle BAC$

$\angle BAC=\frac{\angle BOC}{2}$
$\angle BAC=\frac{160}{2}={80}^{\circ }$
$\therefore \angle BAC={80}^{\circ }$